2 g of metal carbonate is neutralized comple | vedclass

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(D) The number of gram equivalents of $HCl$ used is calculated as: $	ext{Equivalents} = 	ext{Normality} 	imes 	ext{Volume (in L)} = 0.1 \ N 	imes

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$200$

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(D) The number of gram equivalents of $HCl$ used is calculated as: $	ext{Equivalents} = 	ext{Normality} 	imes 	ext{Volume (in L)} = 0.1 \ N 	imes 0.1 \ L = 0.01 \ 	ext{eq}$.According to the law of equivalence, the number of gram equivalents of the metal carbonate must be equal to the number of gram equivalents of $HCl$ used for neutralization.Therefore, the number of gram equivalents of metal carbonate $= 0.01 \ 	ext{eq}$.The equivalent weight is defined as the mass of the substance divided by the number of gram equivalents.$	ext{Equivalent weight} = \frac{	ext{Mass}}{	ext{Number of gram equivalents}} = \frac{2 \ g}{0.01 \ 	ext{eq}} = 200 \ g/	ext{eq}$.Thus, the equivalent weight of the metal carbonate is $200$.

Isotope	Isotopic molar mass $(g \, mol^{-1})$	Abundance (%)
$^{36}Ar$	$35.96755$	$0.337$
$^{38}Ar$	$37.96272$	$0.063$
$^{40}Ar$	$39.9624$	$99.600$

$2 \ g$ of metal carbonate is neutralized completely by $100 \ mL$ of $0.1 \ N$ $HCl$. The equivalent weight of metal carbonate is:

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Isotope Isotopic molar mass $(g \, mol^{-1})$ Abundance (%)

$^{36}Ar$ $35.96755$ $0.337$

$^{38}Ar$ $37.96272$ $0.063$

$^{40}Ar$ $39.9624$ $99.600$

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